POJ 1087 A Plug for UNIX

链接：http://poj.org/problem?id=1087

题意：有n（1≤n≤100）个插座，每个插座用一个数字字母式字符串描述（至多有24 个字符）。有m（1≤m≤100）个设备，每个设备有名称，以及它使用的插头的名称；插头的名称跟它所使用的插座的名称是一样的；设备名称是一个至多包含24 个字母数字式字符的字符串；任何两个设备的名称都不同；有k（1≤k≤100）个转换器，每个转换器能将插座转换成插头。

样例：

4
A
B
C
D
5
laptop B
phone C
pager B
clock B
comb X
3
B X
X A
X D

思路：建立一个汇点，每个插座和汇点连边，流量代表该种插座的个数。建立一个源点，源点和每个设备连边，流量为1。每个转换器将两个插座连边，流量为INF。然后求最大流。答案就是设备的个数减去最大流。

细节：这道题在建图的时候比较恶心。需要注意的是，在m个设备和插座以及k个转换器的数据中，可能有之前没有出现过的设备。所以必须将插座的信息存起来，可以用map来存插座对应的点。

代码：

/*
ID: [email protected]
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
//#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 610;
const int maxm = 2000000;
int m, k;
struct node {
    int v;    // vertex
    int cap;    // capacity
    int flow;   // current flow in this arc
    int nxt;
} e[maxm * 2];
int g[maxn], cnt;
int st, ed, n;
void add(int u, int v, int c) {
    e[++cnt].v = v;
    e[cnt].cap = c;
    e[cnt].flow = 0;
    e[cnt].nxt = g[u];
    g[u] = cnt;

    e[++cnt].v = u;
    e[cnt].cap = 0;
    e[cnt].flow = 0;
    e[cnt].nxt = g[v];
    g[v] = cnt;
}
map<string, int> mp;
map<string, int> :: iterator it;
void init() {
    mem(g, 0);
    cnt = 1;
    mp.clear();
    scanf("%d", &n);
    char str[25];
    ed = 0;
    int tot = 1, has[111] = {0};
    for (int i = 1; i <= n; i++) {
        scanf("%s", str);
        string ss(str);
        int now;
        if (mp[ss] == 0) {
            mp[ss] = tot;
            now = tot++;
        } else now = mp[ss];
        has[now]++;
    }
    for (int i = 1; i < tot; i++) add(i, ed, has[i]);
    scanf("%d", &m);
    char s[25];
    vector<int> v;
    for (int i = 0; i < m; i++) {
        scanf("%s%s", str, s);
        v.pb(tot);
        if (mp[s]) add(tot++, mp[s], 1);
        else {
            mp[s] = tot + 1;
            add(tot, tot + 1, 1);
            tot += 2;
        }
    }
    scanf("%d", &k);
    for (int i = 0; i < k; i++) {
        scanf("%s%s", str, s);
        if (mp[str] == 0) mp[str] = tot++;
        if (mp[s] == 0) mp[s] = tot++;
        add(mp[str], mp[s], INF);
    }
    st = tot;
    for (int i = 0; i < v.size(); i++) add(st, v[i], 1);
    n = tot + 3;
}

int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
    int font = 0, rear = 1;
    for (int i = 0; i <= n; i++) { //n为总点数
        dist[i] = maxn;
        numbs[i] = 0;
    }
    q[font] = ed;
    dist[ed] = 0;
    numbs[0] = 1;
    while(font != rear) {
        int u = q[font++];
        for (int i = g[u]; i; i = e[i].nxt) {
            if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
            dist[e[i].v] = dist[u] + 1;
            ++numbs[dist[e[i].v]];
            q[rear++] = e[i].v;
        }
    }
}
int maxflow() {
    rev_bfs();
    int u, totalflow = 0;
    int curg[maxn], revpath[maxn];
    for(int i = 0; i <= n; ++i) curg[i] = g[i];
    u = st;
    while(dist[st] < n) {
        if(u == ed) {   // find an augmenting path
            int augflow = INF;
            for(int i = st; i != ed; i = e[curg[i]].v)
                augflow = min(augflow, e[curg[i]].cap);
            for(int i = st; i != ed; i = e[curg[i]].v) {
                e[curg[i]].cap -= augflow;
                e[curg[i] ^ 1].cap += augflow;
                e[curg[i]].flow += augflow;
                e[curg[i] ^ 1].flow -= augflow;
            }
            totalflow += augflow;
            u = st;
        }
        int i;
        for(i = curg[u]; i; i = e[i].nxt)
            if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
        if(i) {   // find an admissible arc, then Advance
            curg[u] = i;
            revpath[e[i].v] = i ^ 1;
            u = e[i].v;
        } else {    // no admissible arc, then relabel this vertex
            if(0 == (--numbs[dist[u]])) break;    // GAP cut, Important!
            curg[u] = g[u];
            int mindist = n;
            for(int j = g[u]; j; j = e[j].nxt)
                if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
            dist[u] = mindist + 1;
            ++numbs[dist[u]];
            if(u != st)
                u = e[revpath[u]].v;    // Backtrack
        }
    }
    return totalflow;
}

int main () {
    init();
    printf("%d\n", m - maxflow());
    return 0;
}